nLab tensor product of modules

Contents

Context

Monoidal categories

monoidal categories

With braiding

With duals for objects

With duals for morphisms

With traces

Closed structure

Special sorts of products

Semisimplicity

Morphisms

Internal monoids

Examples

Theorems

In higher category theory

Contents

Idea

The tensor product of modules.

Definition

Definition

Let $R$ be a commutative ring and consider the multicategory $R$Mod of $R$-modules and $R$-multilinear maps. In this case the tensor product of modules $A\otimes_R B$ of $R$-modules $A$ and $B$ can be constructed as the quotient of the tensor product of abelian groups $A\otimes B$ underlying them by the action of $R$; that is,

$A\otimes_R B = A\otimes B / (a,r\cdot b) \sim (a\cdot r,b).$
Definition

The tensor product $A \otimes_R B$ is the coequalizer of the two maps

$A\otimes R \otimes B \;\rightrightarrows\; A\otimes B$

given by the action of $R$ on $A$ and on $B$.

Remark

If the ring $R$ happens to be a field, then $R$-modules are vector spaces and the tensor product of $R$-modules becomes the tensor product of vector spaces.

Remark

This tensor product can be generalized to the case when $R$ is not commutative, as long as $A$ is a right $R$-module and $B$ is a left $R$-module. More generally yet, if $R$ is a monoid in any monoidal category (a ring being a monoid in Ab with its tensor product), we can define the tensor product of a left and a right $R$-module in an analogous way. If $R$ is a commutative monoid in a symmetric monoidal category, so that left and right $R$-modules coincide, then $A\otimes_R B$ is again an $R$-module, while if $R$ is not commutative then $A\otimes_R B$ will no longer be an $R$-module of any sort.

Remark

The tensor product of modules can be generalized to the tensor product of functors.

Properties

Monoidal category structure

The category $R$Mod equipped with the tensor product of modules $\otimes_R$ becomes a monoidal category, in fact a distributive monoidal category.

Proposition

A monoid in $(R Mod, \otimes)$ is equivalently an $R$-algebra.

Proposition

The tensor product of modules distributes over the direct sum of modules:

$A \otimes \left(\oplus_{s \in S} B_s\right) \simeq \oplus_{s \in S} ( A \otimes B_c ) \,.$

Exactness properties

Let $R$ be a commutative ring.

Proposition

For $N \in R Mod$ a module, the functor of tensoring with this module

$(-) \otimes_R N \colon R Mod \to R Mod$
Proof

The functor is additive by the distributivity of tensor products over direct sums, prop. .

A general abstract way of seeing that the functor is right exact is to notice that $(-)\otimes_R N$ is a left adjoint functor, its right adjoint being the internal hom $[N,-]$ (see at Mod). By the discussion at adjoint functor this means that $(-) \otimes_R N$ even preserves all colimits, in particular the finite colimits.

Remark

The functor $(-)\otimes_R N$ is not a left exact functor (hence not an exact functor) for all choices of $R$ and $N$.

Example

Let $R \coloneqq \mathbb{Z}$, hence $R Mod \simeq$ Ab and let $N \coloneqq \mathbb{Z}/2\mathbb{Z}$ the cyclic group or order 2. Moreover, consider the inclusion $\mathbb{Z} \stackrel{\cdot 2}{\hookrightarrow} \mathbb{T}$ sitting in the short exact sequence

$0 \to \mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0 \,.$

The functor $(-) \otimes \mathbb{Z}/2\mathbb{Z}$ sends this to

$0 \to \mathbb{Z}/2\mathbb{Z} \stackrel{0}{\to} \mathbb{Z}/2\mathbb{Z} \stackrel{id}{\to} \mathbb{Z}/2\mathbb{Z} \to 0 \,.$

Here the morphism on the left is the 0-morphism: in components it is given for all $n_1, n_2 \in \mathbb{Z}$ by

\begin{aligned} (n_1, n_2 mod 2) & \mapsto (2 n_1, n_2 mod 2) \\ & \simeq 2 (n_1, n_2 mod 2) \\ & \simeq (n_1, 2 n_2 mod 2) \\ & \simeq (n_1, 0) \\ & \simeq 0 \end{aligned} \,.

Hence this is not a short exact sequence anymore.

One kind of module $N$ for which $(-)\otimes_R N$ is always exact are free modules.

Example

Let $i \colon N_1 \hookrightarrow N_2$ be an inclusion of a submodule. For $S \in$ Set write $R^{\oplus {\vert S\vert}} = R[S]$ for the free module on $S$. Then

$i \otimes_R N \colon N_1 \otimes_R R^{\oplus {\vert S\vert}} \to N_2 \otimes_R R^{\oplus {\vert S\vert}}$

is again a monomorphism. Indeed, due to the distributivity of the tensor product over the direct sum and using that $R \in R Mod$ is the tensor unit, this is

$i^{\oplus {\vert S\vert}} \colon N_1^{\oplus {\vert S\vert}} \hookrightarrow N_2^{\oplus {\vert S\vert}} \,.$

There are more modules $N$ than the free ones for which $(-)\otimes_R N$ is exact. One says

Definition

If $N \in R Mod$ is such that $(-)\otimes_R N \colon R Mod \to R Mod$ is a left exact functor (hence an exact functor), $N$ is called a flat module.

Remark

For a general module, a measure of the failure of $(-)\otimes_R N$ to be exact is given by the Tor-functor $Tor^1(-,N)$. See there for more details.

References

Textbook accounts:

Lecture notes: